NOTE: your equation is not fully balanced; placing a #2# in front of the #"Na"_3"PO"_4# species will do it.
The first thing you'll want to do in any molarity problem like this is to use the given molarity and volume of one substance to calculate the number of moles of that substance.
To do this, we first need to convert the #"CuCl"_2# volume to liters:
#59.4cancel("mL CuCl"_2)((1"L CuCl"_2)/(10^3cancel("mL CuCl"_2))) = 0.0594# #"L CuCl"_2#
Now let's calculate the number of moles of #"CuCl"_2#:
#"mol CuCl"_2 = ("molarity")("L CuCl"_2)#
#= (0.102"mol"/(cancel("L")))(0.0594cancel("L")) = color(red)(0.00606# #color(red)("mol CuCl"_2#
Now that we have the moles of #"CuCl"_2#, let's use the coefficients of the chemical equation to calculate the relative number of moles of #"Na"_3"PO"_4#:
#color(red)(0.00606# #cancel(color(red)("mol CuCl"_2))((2"mol Na"_3"PO"_4)/(3cancel("mol CuCl"_2))) = 0.00404# #"mol Na"_3"PO"_4#
Finally, we'll use this value and given molarity of the #"Na"_3"PO"_4# solution to calculate the volume of the solution:
#"L soln" = "mol solute"/"molarity" = (0.00404cancel("mol Na"_3"PO"_4))/(0.165cancel("mol")/"L")#
#= 0.0245# #"L Na"_3"PO"_4# #"soln" = color(blue)(24.5# #color(blue)("mL Na"_3"PO"_4# #color(blue)("soln"#
Therefore, in order to react completely with #59.4# #"mL"# of a #0.102M# #"CuCl"_2# solution, you must prepare #color(blue)(24.5# #color(blue)("mL"# of a #0.165M# solution of #"Na"_3"PO"_4#.
